JEE Main Mathematics - Set Theory Relations and Mappings
Exam Duration: 60 Mins Total Questions : 30
Given A={2,B,3} where B={4,5} which of the following statements is true?
- (a)
\(4\in A\)
- (b)
\(5\in A\)
- (c)
\(\{ 4,5\} \in A\)
- (d)
\( 4,5\ \in A\)
The element of the set is {4, 5} and not 4 or 5.
n(A)=20,n(B)=10,n(\(A\cup B\))=25, then \(n(A\cap B)\)equals
- (a)
55
- (b)
30
- (c)
5
- (d)
none of these
We know
n(A\(\cup\)B) = n(A) + n(B) - n(A\(\cap\)B)
= 20 + 10 - 25 = 5.
If \(f:R\rightarrow R,f(x)={ x }^{ 2 }\),then f is
- (a)
injective but not surjective
- (b)
surjective but not injective
- (c)
bijective
- (d)
none of these
f (x) = x2, for x = \(\pm\)1, f(x) = 1, i.e., image of x = ± I is the same element 1 in B; so the function is surjective but not injective.
Let the set \(A=\left\{ x\epsilon Z|0\le x\le 10 \right\} \) and R be the relation defined by R={(a,b)|a=b}. Then, R is
- (a)
reflexive and symmetric
- (b)
symmetric and transitive
- (c)
reflexive and transitive
- (d)
equivalence
The given set, A = {x \(\in\) Z I 0 \(\le\) x \(\le\) 10} and the relation R is
defined by R = {(a, b) l a = b}
Now, let a \(\in\) A, then a = a ⇒ (a, a) \(\in\) R
Thus, R is reflexive.
Let a, b \(\in\) A such that (a, b) \(\in\) R, then
a = b ⇒ b = a ⇒ (b, a) \(\in\)R
Thus, R is symmetric.
Now, let a, b, c \(\in\) A such that (a, b) and (b, c) \(\in\) R, then
a = b and b = c ⇒ a = c ⇒ (a, c) \(\in\) R
Thus, R is transitive also.
Hence, R is an equivalence relation.
Consider the following statements
\(I.A\cap B=\phi ,then\quad either\\ \quad \quad \quad \quad A=\phi orB=\phi \\ II.For\quad a\neq b,\left\{ a,b \right\} =\left\{ b,a \right\} \\ \quad \quad \quad \quad \quad and(a,b)\ne(b,a)\\ III.If\quad A\subseteq B,then\\ \quad \quad \quad \quad \quad A\times A\subseteq (A\times B)⋂(B\times A)\\ IV.If\quad A\subseteq B,then\quad (A\times C)\subseteq B\times D\)
Which of these is correct?
- (a)
I, II and III
- (b)
Both I and IV
- (c)
only IV
- (d)
II, III and IV
If A \(\cap\) B = \(\phi\), then it is not necessary that A=ф or B=ф
II. It is true {a,b}={b,a} and (a,b)≠(b,a)
III. By properties of cartesian product
If A⊆B, then AxA⊆(AxB)⋂(BxA)
IV. If A⊆B and C⊆D, then AxC⊆BxD
The set builder form of given set A= {3, 6, 9, 12} and B={1,4,9,..., 100} is
- (a)
A={x:x=3n, n∈N and 1 < n < 5}
B={x:x =n2, n∈N and 1 < n < 10} - (b)
A={x:x=3n, n∈N and 1 < n < 4}
B={x:x =n2, n∈N and 1 < n < 10} - (c)
A={x:x=n2, n∈N and 1 < n < 4}
B={x:x =3n, n∈N and 1 < n < 10} - (d)
None of these
Given, A={3,6,9, 12}
= {x:x = 3n, n∈N and 1 < n < 4}
and B={1,4,9, ..., 100} = {x:x =n2, n∈N and 1 < n < 10}
The number of elements in the set {(a,b):2a2+3b2=35, a,b∈Z}, where Z is the set of all integers, is
- (a)
2
- (b)
4
- (c)
8
- (d)
12
2a2+3b2=35, a,b∈Z
If a=2, b=3, then 2x22+3x32=35
ஃ (a,b) is (2,3). In the same way, other sets are (-2,-3), (4,1), (-4,-1), (2,-3), (-2,3), (-4,1), (4,-1)
So, number of elements in the set is 8.
Which of the following sets is a finite set?
- (a)
A = {x:x ∈ Z and x2-5x+6=0}
- (b)
B={x:x ∈Z and x2 is even}
- (c)
D = {x:x ∈ Z and x > -10}
- (d)
All of these
(a) A = {x:x ∈ Z and x2-5x+6=0}={2,3}
So, A is a finite set
(b)={x:x∈Z and x2is even}={..., -4, -2, 0, 2, 4, ...}
Clearly, B is an infinite set
(c) D = {x:x∈Z and >-10}={-9, -8, -7, ...}
Clearly, D is infinite se
(AxB)∩(AxC)=
- (a)
{(1,4), (2,4), (3,4)}
- (b)
{(4,2), (3,4), (4,1)}
- (c)
{(2,4), (1,4)}
- (d)
None of these
AxB={(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)} and AxC={(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)}
Therefore, (AxB)⋂(AxC)={(1,4), (2,4), (3,4)}
Ax(B∩C)=
- (a)
{(1,3), (1,4), (1,5), (2,5), (2,4), (3,3), 93,6), (3,4)}
- (b)
{(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)}
- (c)
{(2,4), (3,3), (3,4), (3,5), (3,6)}
- (d)
None of these
We have BUC={3,4,5,6}
ஃ Ax(BUC)={(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)}
Let A={x,y,z} and B={a,b,c,d}. Which one of the following is not a relation from A to B?
- (a)
{(x,a), (x,c)}
- (b)
{(y,c}, (y,d)}
- (c)
{(z,a), (z,d)}
- (d)
{(z,b), (y,b), (a,d)}
R⊆ AxB
For given A={x,y,z} & B={a,b,c,d}
AxB={(x,a), (x,b), (x,c), (x,d), (y,a), (y,b), (y,c), (y,d), (z,a), (z,b), (z,c), (z,d)}
Clearly {(z,b), (y,b), (a,d) is not the subset of AxB.
ஃ It is not a relation.
The relation R defined on set A={x:|x|<3, x∈I} by R={(x,y):y=|x|} is
- (a)
{(-2,2), (-1,1), (0,0), (1,1), (2,2)}
- (b)
{(-2,-2), (-2,2), (-1,1), (0,0), (1,-2), (1,2), (2,-1), (2,-2)}
- (c)
{(0,0), (1,1), (2,2)}
- (d)
None of these
Given, A={x:|x|<3, x∈I}
A={x:-3<x<3, x∈I}={-2,-1,0,1,2}
Also, R={(x,y):y=|x|}
ஃ R={(-2,2), (-1,1), (0,0), (1,1), (2,2)}
The range of the function f(x)=\(\frac{x-2}{2-x}\) is
- (a)
R
- (b)
R-{1}
- (c)
{-1}
- (d)
R-{-1}
f(x)=\(\frac{x-2}{2-x}=-\frac{x-2}{x-2}=-1, if x≠2\)
ஃ Range of f={-1}
If A and B be any two sets, then A∩(AUB)' is equal to
- (a)
A
- (b)
B
- (c)
Φ
- (d)
None of these
A∩(AUB)' = A∩(A'∩B') = (A∩A')∩B'
=Φ∩B'=Φ
The domain of the function f(x)=\(\frac{1}{\sqrt{9-x^2}}\) is
- (a)
-3≤x≤3
- (b)
-3<x<3
- (c)
-9≤x≤9
- (d)
-9<x<9
f(x)=\(\frac{1}{\sqrt{9-x^2}}\)
Clearly, 9-x2>0 ⇒ x2-9<0 ⇒ (x+3)(x-3)<0
Thus domain of f(x) is x∈(-3,3).
If n(A)=8and n(A∩B)=2, then n((A∩B)'∩A) is equal to
- (a)
2
- (b)
4
- (c)
6
- (d)
8
n[(A∩B)'∩A]
=n[A'∩B')∩A]
=n[(A'∩A)U(B'∩A)]
=n(Φ(A∩B'))=n(A∩B')
=n(A-B)=n(A)-n(A∩B)=8-2=6
The domain of the function f given by f(x)=\(\frac{1}{\sqrt{x-|x|}}\) is
- (a)
R
- (b)
R+
- (c)
R-
- (d)
{ф}
Given that f(x)=\(\frac{1}{\sqrt{x-|x|}}\)
where x-|x|=\(\begin{cases} x-x=0,\quad if\quad x\ge 0 \\ x-\left( -x \right) =2x,\quad if\quad x<0 \end{cases}\)
Thus \(\frac{1}{\sqrt{x-|x|}}\) is not defined for any x∈R.
Hence f is not defined for any x∈R.
i.e., domain of f={ф}
In a certain town, 25% families own a phone and 15% own a car, 65% families own neither a phone nor a car, 2000 families own both a car and a phone.
Consider the following statements in this regard:
1. 10% families own both a car and a phone
2. 35% families own both either a car or a phone
3. 40,000 families live the town
Which of the above statements are correct?
- (a)
1 and 2
- (b)
1 and 3
- (c)
3 and 3
- (d)
1, 2 and 3
Let X, P and C denote the sets of all families, families owning phone an families owning car respectively. Let total number of families be k.
∴ n(P)=25% of \(k={k\over 4}\), n(C)=15% of \(k={3k\over 20}\)
\(n(\bar P∩\bar C)=65\% \) of \(k={13k\over 20},\)n(P∩C)=2,000
Now \(n(\bar P∩\bar C)=n(X)-n(PUC)⇒ n(PUC)={7k\over20}\)
Also, n(PUC)=n(P)+n(C)-n(P∩C)
\(⇒{7k\over 20}={k\over 4}+{3k\over 20}-n(P∩C)\)
\(⇒n(P∩C)={k\over 20}⇒2,000={k\over 20}⇒k=40,000\)
∴ Statement '1' is incorrect.
\(⇒n(PUC)={7k\over 20}={7\times40,000\over20}=14,000\)
\(={14,000\over 40,000}\times 100\%=35\%\)
∴ Statement '2' is correct
From 50 student taking examinations in Mathematics, Physics and Chemistry, 37 passed Mathematics, 21 Physics and 43Chemistr. At most 29 passed Mathematics and Chemistry and at most 20 passed Physics and Chemistry. The largest possible number that could have passed three examinations is
- (a)
11
- (b)
12
- (c)
13
- (d)
14
Let M, P and C be the sets of students taking examination in Mathematics, Physics and Chemistry respectively.
∴ n(MUPUC)=50, 50, n(M)=37, n(P)=24, n(C)=43, n(M∩P) < 19, n(M∩C) < 29, n(P∩C) < 20.
We have
n(MUPUC)=n(M)+n(P)+n(C)-n(M∩P)-n(M∩C)-n(P∩C)+n(M∩P∩C)
⇒ 50 = 37+24+43-n(M∩P)-n(M∩C)-n(P∩C)+n(M∩P∩C)
⇒ n(M∩P∩C)=n(M∩P)+n(M∩C)+n(P∩C)-54
⇒ n(M∩P∩C) = < 19-29+20-54=14
⇒ n(M∩P∩C) = < 14
Given n(U)=20, n(A)=12,n(B)=9, n(A∩B)=4, where U is the universal set, A and B are subsets of U, then n[(AUB)c] equals to
- (a)
17
- (b)
9
- (c)
11
- (d)
3
n[(AUB)c]=n(U)-n(AUB)
=20-(n(A)+n(B)-n(A∩B))
=20-(12+9-4)=20-17=3
If n(A)=1000, n(B)=500, n(A∩B) > 1 and n(AUB=p, then
- (a)
500 < p < 1000
- (b)
1001< p< 1498
- (c)
100< p< 64
- (d)
1000< p< 1499
We know that, n(AUB)=n(A)+n(B)-n(A∩B)
⇒ p=1000+500-n(A∩B) ⇒1 < n(A∩B) < 500
Hence, p=1000+500-1=1499
an p=1000+500-500=1000
∴ 1000 < p < 1499
State T for true and F for False.
(i) For any real number, √x2=|x|.
(ii) [x+n]=[x]+n, n is an integer and x is any real number between n and n+1
(iii) Domain of irrational function varies from function to function.
(iv) If a and b are positive real numbers, and x is any real number, then a2≤x2≤b2 ⇔ a≤|x|≤b ⇔ x∈(-b,-a]U[a,b]
- (a)
(i) (ii) (iii) (iv) T T F T - (b)
(i) (ii) (iii) (iv) T F T T - (c)
(i) (ii) (iii) (iv) T T T T - (d)
(i) (ii) (iii) (iv) T T T F
(i) True (ii) True (iii) True (iv) False
If a and b are positive real numbers and x is any real number, then a2≤x2≤b2 ⇔ a≤|x|≤b ⇔ x∈(-b,-a]U[a,b]
The domain of the function f defined by f(x)=\(\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}\) is equal to
- (a)
(-∞,-1)⋃(1,4]
- (b)
(-∞,-1]⋃(1,4]
- (c)
(-∞,-1)⋃[1,4]
- (d)
(-∞,-1)⋃[1,4)
Given, \(f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}\)
f(x) is defined when, 4-x≥0 and x2-1>0 ⇒ x≤4 and x2>1
⇒ x≤4 and x∈(-∞,-1)⋃(1,∞)
ஃ x∈(∞,-1)⋃(1,4]
The domain and range of the function f given by f(x)=2-[x-5] is
- (a)
Domain=R+, Range=(-∞,1]
- (b)
Domain=R, Range=(-∞,2]
- (c)
Domain=R, Range=(-∞,2)
- (d)
Domain=R+, Range=(-∞,2]
Given, f(x)=2-|x-5|
Domain of f(x) is defined for all real values of x.
Since, |x-5|≥0 ⇒ -|x-5|≤0
⇒ 2-|x-5|≤2 ⇒ f(x)≤2
Hence, range of f(x) is (-∞,2].
Statement I: The range of the function f(x)=|x| is [0,∞).
Statement II: \(\left| x \right| =\begin{cases} x:\quad x\ge 0 \\ -x:x<0 \end{cases}\)
- (a)
If both Statement I and Statement II are true and Statement II is the correct explanation of Statement I
- (b)
If both Statement I and Statement II are true but Statement II is not the correct explanation of Statement I.
- (c)
If Statement I is true but Statement II is false
- (d)
If Statement I is false and Statement II is true.
Since Statement II is true.
ஃ R(f)=[0,∞)
In a school, there are 20 teachers who teach mathematics or physics. Of these, 12 teach mathematics and 4 teachers teach both mathematics and physics. Then the number of teachers teaching only physics are
- (a)
12
- (b)
16
- (c)
8
- (d)
4
Let M= set of mathematics teachers and P = set of physics teachers.
n(only mathematics teacher) = n(M) - n(M⋂P)=12-4=8
Also, n(MUP)=n(only mathematics teachers)+n(M⋂P)+n (only physics teachers)
ஃ 20 = 8 + 4 + n (only physics teachers)
ஃ n(only physics teachers) = 20 - 12 = 8
If A and B are respectively the sets having the elements as the zeros of the polynomials x3-4x2+x+6 and x3-6x2+11x-6, then match the following.
| Column I | Column II |
|---|---|
| (i) A-B | (p) {-1, 2, 3} |
| (ii) B - A | (q) {-1} |
| (iii) A-(B-A) | (r) {1} |
- (a)
(i) ⟶ (q), (ii) ⟶ (p), (iii) ⟶ (r)
- (b)
(i) ⟶ (q), (ii) ⟶ (r), (iii) ⟶ (p)
- (c)
(i) ⟶ (p), (ii) ⟶ (r), (iii) ⟶ (q)
- (d)
(i) ⟶ (r), (ii) ⟶ (p), (iii) ⟶ (q)
Let f(x) =.0 - 4x2 + x + 6
Zeroes of f(x) = -1, 2, 3
ஃ A = {-1, 2, 3}
Also, let g(x) =x3 - 6x2 + 11x - 6
Zeroes of g(x) = 1,2,3
ஃ B = {1, 2, 3}
(i) A-B={-1} (ii) B-A={1}
(iii) A-(B-A)={-1,2,3}
If sets A and B are defined as A={(x,y)|y=\(\frac{1}{x}\), 0≠x∈R}
B={(x,y)|y=-x,x∈R}, then
- (a)
A⋂B=A
- (b)
A⋂B=B
- (c)
A⋂B=ф
- (d)
AUB=A
Since -x≠\(1\over x\)
ஃ There is no common element in A and B.
Statement-I: If A = {x: x is a multiple of 3} and B = {x: x is a multiple of 5} then A⋂B={x|x is a multiple of 15}
Statement-II: A⋂B is a set which contain both the elements of A and B.
- (a)
If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I.
- (b)
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I.
- (c)
If Statement-I is true but Statement-II is false.
- (d)
If Statement -I is false and Statement -II is true.
Since x∈A⋂B ⇔ x∈A and x∈B ⇔ x is a multiple of 3 and x is a multiple of 5 ⇔ x is a multiple of 15.
Hence A∩B={x|x is a multiple of 15}={15,30,45,...}
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